It was about dimensionlessness of bulk scalars from the perpective of boundary conformal transformation. If we have a bulk scalar $\phi$ it is dimensionless from the point of boundary transformations as (as answered by Motl)

``it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. ''

But the way we talk about conformal dimension of a boundary operator is through it's transformation under the diffeomorphism $x\to\lambda x$ and then if a boundary field $f(x)\to f^\prime(x)=\lambda^hf(\lambda x)$ (or the relation written later by Motl in that answer), then $h$ is it's conformal dimension. I hope what I said so far is right. But, isn't conformal transformation at the boundary also act as a diffeomorphism in the bulk (at least for boundary part of the coordinates of the bulk field). Then why will it not transform in such a way that it has a dimension?

I believe, e.g. spin-1 gauge field in the bulk has (confomal/scaling/mass) dimension 1. So, how does it have a dimension? Sorry for the novice question and thanks in advance!

Hi, as I said previously, and you even quoted me again, yes, the scalings on the boundaries are dual to diffeomorphisms in the bulk but these are special diffeomorphisms, namely isometries, i.e. diffeomorphism that preserve the metric. So the bulk field only shifts to another point, it doesn't scale in any way, so there is no way to associate any dimension with it. The usual mass dimension of the bulk fields - incidentally, in $AdS_5$, the dimension of bosonic fields is $m^{3/2}$, not $m$ - depends on conventions and has no canonical interpretation on the CFT side.
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Luboš MotlFeb 25 '13 at 7:08

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The reason why the usual mass dimension of bulk fields isn't physically important is exactly that the theory in the bulk is not scale-invariant or conformal. That means that the fields in it don't really have well-defined dimensions or at least these dimensions would depend on the renormalization scale. Alternatively, one could try to consider the UV or IR fixed points - limits - but the UV fixed point doesn't exist (or at least is unknown) in gravitational theories and the IR limit is modified by the AdS-sized curvature.
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Luboš MotlFeb 25 '13 at 7:10

Please stay with me for few more stages of confusion! I thought the way fields transform under dilatations (namely my previously written transformation for $f(x)$ in this question) is a special case of or another example of active transformation of fields. And I thought active transformation can happen even if it's an isometry (metric preserved). So, $x\mapsto\lambda x$ in the boundary still remains $x\mapsto\lambda x$ in bulk, albeit it's now isometry; but that doesn't not give us any active transformation.
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user1349Feb 25 '13 at 14:46